[斐波那契數列]證明

好多方法可以證明 (例如維基)﹐以下用最簡單的數學歸納法証明
P(n)為命題
斐波那契數的通項是S(n)=1/√5{[(1+√5)/2]^n-[(1-√5)/2]^n}
當n=1
S(1)=1
所以P(1)成立
當n=2
S(2)
=1/√5{[(1+√5)/2]^2-[(1-√5)/2]^2}
=2√5/√5
=1
所以P(2)成立
設P(k),P(k+1)成立
S(k)=1/√5{[(1+√5)/2]^k-[(1-√5)/2]^k}
S(k+1)=1/√5{[(1+√5)/2]^(k+1)-[(1-√5)/2]^(k+1)}
當n=k+2
由S(k+2)=S(k)+S(k+1)
S(k+1)
=1/√5{[(1+√5)/2]^k-[(1-√5)/2]^k}+1/√5{[(1+√5)/2]^(k+1)-[(1-√5)/2]^(k+1)}
=1/√5{[(1+√5)/2]^k[1+[(1+√5)/2]]-[(1-√5)/2]^k[1+[(1-√5)/2]]}
=1/√5{[(1+√5)/2]^k[(3+√5)/2]-[(1-√5)/2]^k[(3-√5)/2]}
=1/√5{[(1+√5)/2]^k[(1+√5)/2]^2-[(1-√5)/2]^k[(1-√5)/2]^2}
=1/√5{[(1+√5)/2]^(k+2)-[(1-√5)/2]^(k+2)}
所以P(k+2)成立
注
[(1+√5)/2]^2=(6+2√5)/4=(3+√5)/2
根據數學歸納法﹐對所有正整數n,P(n)成立

2007-12-12 20:07:41 補充：
打錯了第18行是S(k+2)